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$u$ and $v$ are co prime positive integers. What is the value of $u+v$? This seems like an easy problem, but I can't figure out what I'm doing wrong.

$f(x)=3x^2+4x+9$

$f(x)=(x-a)(x-b)$

$f(-x)=(x+a)(x+b)$

$f(-1)=(1+a)(1+b)$

$f(-1)=3-4+9=8$

$\large \frac 81 =\large \frac uv$

$u+v=9$

The answer is supposed to be $11$. What am I doing wrong? Thanks.

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    $\begingroup$ Note: $f(x)=\color{red}{3}(x-a)(x-b)$, not $(x-a)(x-b)$. $\endgroup$
    – user1551
    Commented Jun 27, 2013 at 18:02
  • $\begingroup$ Ahh ok I see thank you $\endgroup$
    – Ovi
    Commented Jun 27, 2013 at 18:06
  • $\begingroup$ @Ovi, I'm interested in improving the Brilliant interface to help students figure out their mistakes. Do you mind sending me an email to get started? $\endgroup$
    – Calvin Lin
    Commented Jun 27, 2013 at 18:50

2 Answers 2

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The formula $f(x)=(x-a)(x-b)$ is not correct. It should be $f(x)=3(x-a)(x-b)$. Then it works. ;)

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Hint: Your product is $1+(a+b)+ab$. The sum of the roots is $\frac{-4}{3}$ and the product of the roots is $\frac{9}{3}$.

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